Compton effect with no scattering angle

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SUMMARY

The discussion centers on calculating the endpoint energy for photons generated when a 0.75 MeV electron beam stops inside a tungsten target. Participants clarify that without a scattering angle, the maximum photon energy is indeed 0.75 MeV, as the equation for Compton scattering simplifies to zero. The conversation also distinguishes between Compton scattering and bremsstrahlung, emphasizing that the scenario described pertains to bremsstrahlung radiation. The key takeaway is that the endpoint energy for the photons is 0.75 MeV.

PREREQUISITES
  • Understanding of Compton scattering and bremsstrahlung radiation
  • Familiarity with photon energy calculations
  • Knowledge of electron interactions with matter
  • Basic principles of quantum mechanics
NEXT STEPS
  • Study the Compton effect and its mathematical derivation
  • Learn about bremsstrahlung radiation and its applications
  • Explore photon energy calculations in different materials
  • Investigate the properties of tungsten as a target material in radiation physics
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Physics students, researchers in particle physics, and professionals working with radiation and photon interactions will benefit from this discussion.

oksanav
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How do I find the endpoint energy for photons when an .75 MeV electron beams stops inside a tungsten target? I just don't understand how to solve it without the scattering angle.:confused:
 
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Here is the exact question, from a notoriously vague teacher:
A .75 MeV electron beam strikes and stops inside a tungsten target.
a. Determine the endpoint energy for the photons.
b. Determine the momentum of the hiest energy photons.
c. Determine the wavelength of the end-point photons.
Wouldn't an angle of 0 make the equation (lamda2-lamda1=(h/mc)(1-cos(0))) go to 0?
 
That's not the Compton effect. It is bremsstrahlung (stopping radiation).
The max photon energy would be .75 MeV.
 
Yeah, I realized he was just trying to get that point across, that the max energy would be .75 MeV. I'm just used to more complicated questions with this guy. Thanks for all the prompt help!
 

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